Proposition 5

Let $A$ be a ring which is not necessarily commutative, $m$ a two-sided ideal of $A$ contained in the Jacobson radical of $A$ and $M$ a left $A$-module. Suppose that one of the following conditions holds: (i) $M$ is finitely presented; (ii) $m$ is nilpotent. Then, if $(A/m) \otimes_A M = M/mM$ is a free left $(A/m)$-module and the canonical homomorphism from $m \otimes_A M$ to $M$ is injective, $M$ is a free $A$-module. To be precise, if $(x_i)_{i \in I}$ is a family of elements of $M$ such that $(1 \otimes x_i)$ is a basis of the $(A/m)$-module $M/mM$, $(x_i)$ is a basis of $M$.

If $a \in A$, $x \in M$ and $\bar{a}$ is the class of $a$ in $A/m$, then $\bar{a} \otimes x = 1 \otimes (ax)$ and hence the hypothesis implies that there exists a family $(x_i)_{i \in I}$ of elements of $M$ such that $(1 \otimes x_i)$ is a basis of the $(A/m)$-module $(A/m) \otimes_A M$. We already know that the $x_i$ generate $M$ (Corollary 2 to Proposition 4); we shall see that they are linearly independent over $A$. To this end, let us consider the free $A$-module $L = A_s^{(I)}$; let $(e_i)$ be its canonical basis and $u : A_s^{(I)} \to M$ the $A$-linear mapping such that $u(e_i) = x_i$ for all $i \in I$; if $R$ is the kernel of $u$, we shall prove that $R = 0$. Under hypothesis (i), $(A/m) \otimes_A M$ is a finitely generated $(A/m)$-module, hence $I$ is necessarily finite and $R$ is a finitely generated $A$-module by Chapter I, § 2, no. 8, Lemma 9. Then by Proposition 4 it will be sufficient to prove (under either hypothesis) that $R = mR$.

Let $j$ be the canonical injection $R \to L$; then there is a commutative diagram

in Which the two rows are exact, $j$ is injective and $1 \otimes u$ is surjective (Chapter

(Chapter I, § 1, no. 4, Proposition 2); it is sufficient to verify that v is bijective, for then we deduce that Coker(a) = 0, in other words that a is surjective and consequently R = mR. Now, Coker(b) = (A/m) \otimes_A L and

and by definition v(1 \otimes e_i) = 1 \otimes x_i; as (1 \otimes e_i) is a basis of (A/m) \otimes_A L, the definition of the x_i shows that v is bijective.